The locus of the points on the curve {y^2 = 4 | vedclass

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(A) Given the curve equation: ${y^2 = 4a(x + a \sin \frac{x}{a})}$.Differentiating with respect to $x$:${2y \frac{dy}{dx} = 4a(1 + a \cos \frac{x}{a}

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$A$ straight line

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(A) Given the curve equation: ${y^2 = 4a(x + a \sin \frac{x}{a})}$.Differentiating with respect to $x$:${2y \frac{dy}{dx} = 4a(1 + a \cos \frac{x}{a} \cdot \frac{1}{a})}$${2y \frac{dy}{dx} = 4a(1 + \cos \frac{x}{a})}$For the tangent to be parallel to the $x$-axis,the slope ${\frac{dy}{dx} = 0}$.This implies ${1 + \cos \frac{x}{a} = 0}$,so ${\cos \frac{x}{a} = -1}$.Therefore,${\frac{x}{a} = (2n+1)\pi}$. For the simplest case,let ${n=0}$,so ${x = a\pi}$.Substituting ${x = a\pi}$ into the original equation:${y^2 = 4a(a\pi + a \sin \pi)}$${y^2 = 4a(a\pi + 0)}$${y^2 = 4a^2\pi}$.Since ${a}$ is a constant,${y^2 = 4a^2\pi}$ represents a set of lines parallel to the $x$-axis (specifically ${y = \pm 2a\sqrt{\pi}}$). Among the given options,this set of horizontal lines is best described as a collection of straight lines.

The locus of the points on the curve ${y^2 = 4a(x + a \sin \frac{x}{a})}$ where the tangent is parallel to the $x$-axis represents . . . .

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